Question

Using the mirror equation and the formula of magnification, deduce that “the virtual image produced by a convex mirror is always diminished in size and is located between the pole and the focus.”

Hint:

Convex mirrors always produce virtual, erect, and diminished images because the reflected rays diverge and appear to come from a point behind the mirror.

Answer 1

Mirror Equation: \[\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\]Convex Mirror Sign Conventions: - Object distance \( u \) is negative as the object is in front of the mirror.- Focal length \( f \) is positive for a convex mirror.- Image distance \( v \) is positive, indicating a virtual image formed on the same side as the object and is upright.Step 1: Analysis with Mirror Equation
With \( f = +f \) and \( u = -|u| \), the equation becomes:\[\frac{1}{v} = \frac{1}{f} - \frac{1}{(-|u|)}\Rightarrow \frac{1}{v} = \frac{1}{f} + \frac{1}{|u|} \Rightarrow v = \left( \frac{1}{f} + \frac{1}{|u|} \right)^{-1}\]Since \( \frac{1}{f} \) and \( \frac{1}{|u|} \) are both positive, \( \frac{1}{v} \) is positive, and \( \frac{1}{v}>\frac{1}{f} \), which implies \( v<f \). Therefore, the image is located between the mirror's pole and its focal point.Step 2: Magnification Formula
\[m = \frac{h'}{h} = \frac{-v}{u}\]Substituting \( v>0 \) and \( u<0 \): \( m = \frac{-v}{- |u|} = \frac{v}{|u|} \)Since \( v<|u| \) (as derived from \( v<f \) and \( |u|>f \) for a convex mirror), the magnification \( m<1 \), meaning the image is reduced in size.Conclusion:
The image formed by a convex mirror is:- Virtual (indicated by \( v>0 \)), - Erect (indicated by \( m>0 \)), - Diminished (indicated by \( m<1 \)), - Located between the pole and the focus (indicated by \( v<f \)).% Final Answer Statement Answer: The mirror equation and magnification formula demonstrate that a convex mirror consistently produces a virtual, erect, and diminished image situated between the pole and the focus.